A quantum channel describes the finite-time evolution of the reduced system of interest. Suppose one has the total unitary of the system and environment taken together, tracing out the environment finally leads to the Kraus operators describing the channel or in other words the dynamical map. Now, the channel/map has in it encoded the effect of environment and so on and it is a subsystem description.

Now, what does one mean when it is said the repeated action of a channel etc? I have seen statements like action of the depolarizing channel many times etc. Since the map is a description for the reduced system, what do these statements mean/how are they implemented?

Since the channel/map is a reduced description, what does it mean by applying a channel n times and so on, since it is not a unitary operator. Something is confusing.


1 Answer 1


Note that a channel must, by definition, map density operators to densityo operators. Now, If the in- and output dimension of a channel $\mathcal{N}$ are the same, there is nothing stopping us from applying the channel twice to a density operator, since applying it the first time gave us a new density operator:

$$ \mathcal{N}(\rho) = \rho',~\mathcal{N}(\mathcal{N}(\rho)) = \mathcal{N}(\rho') = \rho''. $$

We can of course this for any number of channel concatenations.

All of this was on a formal level, i.e. if we have mathematical description of our channel (such as the Kraus representation you mentioned), and the in- and output dimensions are the same, then there is nothing stopping us from defining the $n$-fold concatenation of a channel $\mathcal{N}$.

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    $\begingroup$ From the Stinespring dilation one can view a channel always as a unitary operation action on the state together with some state prepared in $|0>$ and then only given access to part of the system. Of course, one could always just 'throw away' part of the system. If one can perform the state preparation and unitary operation perfectly, then one can implement such a channel in a controlled manner. Normally of course you wouldn't want to apply a (non-unital) channel, but the channel just arises from you having to perform a certain operation in an imperfect manner. $\endgroup$ Commented Jun 13, 2017 at 11:34

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